What is the integral of 1/((x^2+2x)^{1/2)} ?

The integral of 1/((x^2+2x)^{1/2)} is ln|(x^2+2x)^{1/2}+x+1|+C

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integral of 1/((x^2+2x)^{1/2)}

integral

\frac{1}{( x ^{2} + 2 x ) ^{\frac{1}{2}}}

ln (x^{2} + 2 x)^{\frac{1}{2}} + x + 1 + C

Solution steps

\int \frac{1}{( x ^{2} + 2 x ) ^{\frac{1}{2}}} d x

Complete the square

x^{2} + 2 x : (x + 1)^{2} - 1

= \int \frac{1}{( ( x + 1 ) ^{2} - 1 ) ^{\frac{1}{2}}} d x

Apply u-substitution

= \int \frac{1}{( u ^{2} - 1 ) ^{\frac{1}{2}}} d u

Apply Integration By Parts

= \frac{u}{( u ^{2} - 1 ) ^{\frac{1}{2}}} - \int - \frac{u ^{2}}{( u ^{2} - 1 ) ^{\frac{3}{2}}} d u

\int - \frac{u ^{2}}{( u ^{2} - 1 ) ^{\frac{3}{2}}} d u = \frac{u}{( u ^{2} - 1 ) ^{\frac{1}{2}}} - ln (u^{2} - 1)^{\frac{1}{2}} + u

= \frac{u}{( u ^{2} - 1 ) ^{\frac{1}{2}}} - (\frac{u}{( u ^{2} - 1 ) ^{\frac{1}{2}}} - ln (u^{2} - 1)^{\frac{1}{2}} + u)

Substitute back

u = x + 1

= \frac{x + 1}{( ( x + 1 ) ^{2} - 1 ) ^{\frac{1}{2}}} - \frac{x + 1}{( ( x + 1 ) ^{2} - 1 ) ^{\frac{1}{2}}} - ln ((x + 1)^{2} - 1)^{\frac{1}{2}} + x + 1

Simplify

\frac{x + 1}{( ( x + 1 ) ^{2} - 1 ) ^{\frac{1}{2}}} - \frac{x + 1}{( ( x + 1 ) ^{2} - 1 ) ^{\frac{1}{2}}} - ln ((x + 1)^{2} - 1)^{\frac{1}{2}} + x + 1 : ln (x^{2} + 2 x)^{\frac{1}{2}} + x + 1

= ln (x^{2} + 2 x)^{\frac{1}{2}} + x + 1

Add a constant to the solution

= ln (x^{2} + 2 x)^{\frac{1}{2}} + x + 1 + C

What is the integral of 1/((x^2+2x)^{1/2)} ?
The integral of 1/((x^2+2x)^{1/2)} is ln|(x^2+2x)^{1/2}+x+1|+C